Originally Posted by

**GrigOrig99** Can anyone help me confirm if I've solved this correctly?

Many thanks.

Q. $\displaystyle 7^n+4^n+1$ is divisible by 6 for n $\displaystyle \in \mathbb{N}$

**Attempt: **__Step 1:__ For n = 1...

$\displaystyle 7^1+4^1+1=12$, which can be divided by 6.

Therefore, n = 1 is true...

__Step 2:__ For n = k...

Assume $\displaystyle 7^k+4^k+1$ can be divided by 6,

i.e. $\displaystyle 7^k+4^k+1=6Z$, where Z is an integer...**1**

Show that n = k + 1 is true...

i.e. $\displaystyle 7^{k+1}+4^{k+1}+1$ can be divided by 4

$\displaystyle 7^{k+1}+7\cdot4^k-7\cdot4^k+4^{k+1}+1$ => $\displaystyle 7(7^k+4^k)+4^k(7-4)+1$...from **1** above => $\displaystyle 7(6Z-1)+4^k\cdot3+1$ => $\displaystyle 42Z-7+4^k\cdot3+1$ => $\displaystyle 42Z+4^k\cdot3-6$ => $\displaystyle 84Z+4^k\cdot6-12$ => $\displaystyle 6(14Z+4^k-2)$

Thus, assuming n = k, we can say n = k + 1 is true and true for n = 2, 3,... and all n $\displaystyle \in \mathbb{N}_0$